An Upper Bound on the Sum Capacity of the Downlink Multicell Processing with Finite Backhaul Capacity

In this paper, we study upper bounds on the sum capacity of the downlink multicell processing model with finite backhaul capacity for the simple case of 2 base stations and 2 mobile users. It is modelled as a two-user multiple access diamond channel. It consists of a first hop from the central processor to the base stations via orthogonal links of finite capacity, and the second hop from the base stations to the mobile users via a Gaussian interference channel. The converse is derived using the converse tools of the multiple access diamond channel and that of the Gaussian MIMO broadcast channel. Through numerical results, it is shown that our upper bound improves upon the existing upper bound greatly in the medium backhaul capacity range, and as a result, the gap between the upper bounds and the sum rate of the time-sharing of the known achievable schemes is significantly reduced.


I. INTRODUCTION
The multi-cell processing system, as reviewed in [1], has been used to increase the throughput and to cope with the inter-cell interference.The downlink multi-cell processing system, when first considered, consists of different base stations linked to the central processor via backhaul links of unlimited capacity, and therefore, the amount of cooperation among the different base stations is unbounded.This network can be modeled by a MIMO broadcast channel and the sum-rate characterization was found in [2].Later on, due to the impracticality of unlimited T. Yang and W. Kang are with the Information Security Research Center, Southeast University, Nanjing, China (email: {tianyu,wkang}@seu.edu.cn).N. Liu is with the National Mobile Communications Research Laboratory, Southeast University, Nanjing, China (email: nanliu@seu.edu.cn).S. Shamai (Shitz) is with the Department of Electrical Engineering, Technion Israel Institute of Technology, Haifa 32000, Israel (e-mail: sshlomo@ee.technion.ac.il).capacity backhaul links, [3]- [7] studied the problem of finding the capacity region of the downlink multicell processing system when the capacities of the backhaul links are finite, and proposed various achievable schemes to efficiently utilize the finite capacity backhaul links.More specifically, in [3], a compressed dirty-paper coding scheme is proposed, where the base stations are treated as the antennas of the central processor and the dirty-paper coding codewords for each antenna are compressed and transmitted on the backhaul links.The scheme is improved in [4] by allowing the quantization noise of the base stations be correlated.The scheme of reverse compute-and-forward was proposed in [5] where linear precoding is performed at the central processor and the backhaul links are used to transmit linear combinations of the messages over a finite field.Such linear precoding transforms the channel seen at each mobile user into a point-to-point channel where integer-valued interference is eliminated by precoding and the remaining noninteger residual interference is treated as noise.By regarding the network model as a multi-user diamond channel, an achievability scheme is proposed in [6], [7] by combining Marton's achievability for the broadcast channel [8] and the achievability of sending correlated codewords over a multiple access diamond channel [9], [10].
The outer bound on the capacity region for this network is unknown except for the simple cut-set bound [11], which is the minimum of the capacity between the first hop from the central processor to the base stations and that of the second hop from the base stations to the mobile users.When the capacity of the backhaul links are relatively large, the performance of the scheme of compressed dirty-paper coding approaches that of the simple cut-set bound.On the other hand, when the capacity of the backhaul links are relatively small, the scheme of reverse compute-and-forward reaches the simple cut-set bound [6].In the medium capacity region, there is still a relatively large gap between the simple cut-set upper bound and the performance of the time-sharing of the known achievable schemes.So it is unknown how well the proposed achievable schemes are and whether further efforts are needed in proposing better achievable schemes than existing ones for the downlink multicell processing system.
In this paper, we derive a novel upper bound on the sum capacity of the downlink multicell processing network consisting of two base stations and two users.Similar to [6], we regard the network as a 2-user multiple access diamond channel.We first provide a cut-set upper bound using more cuts than the known simple cut-set bound of the minimum between the capacities of the first and the second hop.Next, single-letterization methods for the Gaussian multiple access September 6, 2016 DRAFT diamond channel [12]- [15] is applied to our problem.Finally, we obtain a novel upper bound on the sum capacity utilizing the converse tools of the Gaussian MIMO broadcast channel in [16].The derived upper bound is expressed in terms of the sum capacity of the Gaussian MIMO broadcast channel given input covariance constraint, which has been found in [16]- [21], and thus, is easy to evaluate numerically.
Comparing numerically the proposed upper bound, the simple cut-set upper bound and the sum rate of various achievable schemes for the multicell processing system in terms of the sumrate, we see that our upper bound improves upon the existing simple cut-set upper bound greatly in the medium backhaul capacity range, and as a result, the gap between the upper bounds and the sum rate of the time-sharing of the known achievable schemes is significantly reduced.

II. SYSTEM MODEL
In this paper, we consider the downlink multicell processing system with two base stations and two users.This network model can be seen as the 2-user multiple access diamond channel [6], see Fig.
An (M 1 , M 2 , n, n ) code consists of an encoding function at the source node: two encoding functions at the relay nodes: and two decoding functions at the destination nodes: The average probability of error is defined as Rate pair (R 1 , R 2 ) is said to be achievable if there exists a sequence of (2 nR 1 , 2 nR 2 , n, n ) code such that n → 0 as n → ∞.The capacity of the 2-user multiple access diamond channel is the closure of the set of all achievable rates pairs.
In this paper, we study the Gaussian case, where , and the channel between the two relays and each destination node is a Gaussian multiple access channel, i.e., the received signals at the destination nodes are where X 1 and X 2 are the input signals from Relays 1 and 2, respectively, U 1 , U 2 are two independent zero-mean unit-variance Gaussian random variables that are independent to (X 1 , X 2 ),

III. AN UPPER BOUND ON THE SUM CAPACITY OF THE 2-USER GAUSSIAN MULTIPLE ACCESS DIAMOND CHANNEL
The following of this paper finds an upper bound on the sum capacity of the 2-user Gaussian multiple access diamond channel.
where C MIMO (ρ) denotes the capacity region of the broadcast channel described in ( 1) and ( 2) where is the transmitted signal of the 2 antennas of the transmitter, and Y 1 and Y 2 are the received signals of the single-antenna Receivers 1 and 2, respectively.The input of the transmitter must satisfy a covariance constraint, i.e., The capacity region of the MIMO broadcast channel, i.e., C MIMO (ρ), has been found in [16]- [18].
C sum MIMO (ρ) defined in (3) is the sum capacity of the corresponding MIMO broadcast channel and it has been found in [19]- [21].
Before we introduce the main theorem, let us define the following functions for ρ ∈ and the following variables where sgn(•) is the sign function of •, and the following sets The following is the main result of this paper.
Theorem 1 The sum-rate R 1 +R 2 is achievable for the 2-user Gaussian multiple access diamond channel only if it satisfies for both x = a and x = b.
Proof: The proof is in Appendix A.
In Thoerem 1, ( 4) is proved using the cut-set bound from the four cuts, i.e., Cuts A, B, C and D of Fig. 2, on the sum rate The more difficult part is to prove that when ρ satisfies The existing simple cut-set upper bound on the sum capacity is

IV. NUMERICAL RESULTS
To illustrate the tightness of the derived upper bound in Theorem 1, we plot and compare the existing simple cut-set upper bound on the sum capcity in (6), the new cut-set upper bound of (4), the new upper bound of Theorem 1, and the achievable sum rates of existing schemes for the 2-user Gaussian multiple access diamond channel.
The results are shown in Fig. 3 for the symmetric case of a = b = 0.9, P 1 = P 2 = 10 and The sum rate achieved by the achievable schemes of sending correlated codewords by the relays [6] , the compressed dirty-paper coding allowing correlated quantization noise [4] and the  reverse computer-and-forward scheme [5] are denoted by the solid, circled, and dashed lines, respectively.Furthermore, the sum rate of the time-sharing of all the existing achievable schemes, which is the largest known lower bound for the sum capacity, is denoted by the dot-dashed line.
In the gap between the derived upper bound in Theorem 1, i.e., the diamond line, and the largest known lower bound for the sum capacity, i.e., the dot-dashed line, lies the sum capacity of the 2-user Gaussian multiple access diamond channel for this symmetric case, and as we can see, the gap is not large, which means that the existing achievable schemes perform reasonably well for this scenario.
In the case of a = 0.9, b = −0.9,P 1 = P 2 = 10 and C 1 = C 2 = C, the results are shown in Fig. 4, and similar observations as Fig. 3 follow.

V. CONCLUSION
In this paper, we derive a novel upper bound on the sum capacity of the 2-user Gaussian multiple access diamond channel.This is done by utilizing the converse tools of the multiple  access diamond channel and that of the Gaussian MIMO broadcast channel.Through numerical results, we show that the derived upper bound improves upon the existing simple cut-set upper bound significantly, and as a result, the gap between the lower and upper bounds on the sum capacity is greatly reduced when the capacities of the backhaul links are in the medium range.

APPENDIX A PROOF OF THEOREM 1
For any sequence of (2 nR 1 , 2 nR 2 , n, n ) code, let X n k denote the input of Relay k into the n uses of the channel p(y 1 , y 2 |x 1 , x 2 ), and Y n k denote the corresponding output received at Receiver k, k = 1, 2. Due to the power constraint, we have Define of a random variable Q that is independent of everything else and uniformly distributed on {1, 2, • • • , n}, further define Define the correlation coefficient between X 1 and X 2 as From ( 7) and ( 8), we have Define Based on (9), we have Hence, , and further define K as We can see that Now, based on the four cuts demonstrated in Fig. 2, we have the following cut-set upper bounds on the sum capacity, i.e., R 1 + R 2 : 1) Considering Cut C, we have 2) Considering Cut D, due to (12), we have 3) Considering Cut B, we have two cases: a) For the case of |b| ≤ 1, where (15) follows from the fact that without loss of generality, we consider deterministic encoding at the source node, i.e., (X n 1 , X n 2 ) is a deterministic function of (W 1 , W 2 ), (17) follows from Fano's inequality, (18) follows from the fact that we consider deterministic encoders and the Markov Chain where Ũ n is an i.i.d.sequence of Gaussian random variables with zero mean and variance 1 b 2 − 1, and it is independent of everything else.Note that given a physically degraded version of Y n 1 .Furthermore, note the similarity between which means that we have Thus, from (18), we continue to write as follows while for the simplicity of presentation, we have dropped the 2n n term, where (19) follows from the fact that given X n 2 , Ỹ n 2 is a physically degraded version of Y n 1 .Define auxiliary random variables Thus, we have September 6, 2016 DRAFT where (20) follows from the Markov Chain V i → (X 1i , X 2i ) → Y 1i , ( 21) follows from the definition in (8) and Note that the sum capacity of the degraded broadcast channel where the input of the channel is X 1 given X 2 = x 2 and the outputs of the channel is Y 1 and Ỹ2 , respectively, is given by [11] max Hence, for the particular p(v, x 1 ) as defined by the codebook and ( 23), we have Hence, following from ( 22) and ( 24), we have where (25) follows from the convexity of the log(•) function, (26) follows from the fact that the mean-squared error (MSE) of the optimal Bayes least square (BLS) estimator is smaller than that of the linear least squared (LLS) estimator, and ( follows from ( 9) and (11).
We now proceed to derive another upper bound on R 1 + R 2 which is valid when ρ * satisfies Using Fano's inequality, we have where (34) follows from the Markov chain . Using Fano's inequality, we further have where (36) is because of 2 ) forms a Markov chain.Thus, omitting the n term which will go to zero and n → ∞, from (35) and (36), we have where (37) follows by introducing a sequence of auxiliary random variables Z n and utilizing the fact that The above derivation is true for any Z n .
Next, we perform the single-letterization of (37).To do this, we restrict ourselves to consider Z n that is the output of the following memoryless Gaussian channel with Y n 2 being the input: where U 3 is a Gaussian random variable with zero mean and variance N 3 .Further define Z Z Q .
We single-letterize (37) by single-letterizing each of the following three terms.
where (39) follows from the Markov chain ( and (40) follows from conditioning reduces entropy and the Markov chain Q 2) Based on (38), we have where (41) follows from the Markov chain ( , and further define Note that the auxiliary random variables thus defined satisfy Based on the definition of the random variables in ( 8) and (43), we have Thus, there exists a γ ≥ 0 such that We have where (52) follows from the distribution of (51), (53) follows from the EPI [24, Lemma I] and where in (54), we have used (9), and (55) follows from the definition of ρ * in (10).
and a, b ∈ R are the channel gains from Relay 1 to Destination 2 and Relay 2 to Destination 1, respectively.Without loss of generality, we take a = 0 and b = 0.The case of a = 0 or b = 0 follows from continuity.The transmitted signals at the two relays must satisfy the average power constraints: for any x n k that Relay k sends into the channel, it must satisfy , b, then we have(5), which is strictly tighter than(4).The converse techniques we use to prove this include 1) the bounding of the correlation between the transmitted signals of the two relays via an auxiliary random variable[12]-[15], which was inspired by Ozarow in solving the Gaussian multiple description problem[22]; 2) the single-letterization technique from[23,  page 314, equation (3.34)]; 3) the entropy power inequality (EPI) [24, Lemma I]; and 4) the derivation of the capacity region of the Gaussian MIMO broadcast channel with private messages in [16, Section III.A].